I know that:
two parallel straight lines cut by a transversal form pairs of angles that enjoy particular relationships between them, and that have specific names. A transversal in cutting two parallel lines identifies alternate internal angles, alternate external angles, internal conjugate angles, external conjugate angles and corresponding angles.
I have to prove that for this image that $\angle A''C''B''\cong \angle ACB$.
My proof:
To decompose the weight force or gravitational force, we need to know what angle it forms with the downward slope. So let's look at the figure and notice two right triangles:
- The triangle $\triangle ABC$ the cathetus $d$ and $h$ with hypotenuse $s$. Where $h$ is the height of the slope and $s$ its length, the angle $\angle CAB$ is right, the angle $\angle BCA$ is $\alpha$ and the angle $\angle ABC$ is $\pi/2-\alpha$;
- The triangle $\triangle A''C''B''$ has hypotenuse $\mathbf{F}_g$ and the cathetus $\mathbf{F}_s$ and $\mathbf{F}_d$.
First we see that the angle in $\hat{B}'$ is equal to $\hat{B}$, because they are corresponding angles of the parallel lines passing through $AB$ and $A'B'$ cut by the transversal passing through $BC$.
Then we see that the angle in $\hat{B}''$ is equal to $\hat{B}'$, because they are interior alternate angles of the parallel lines passing through $BC$ and $A''B'$ cut by the line passing through $B'C'$. From the criteria of similarity we know that if two trinagles have two coincident angles then they are similar. From the first criterion of similarity of triangles we know that:
Two triangles are similar if and only if they have a congruent angle and the two sides adjacent to it are proportional to each other according to the same constant $k$.
Since I suspect they do not remember or to find difficult my explanation is there a simple, alternative way to prove that $\angle ACB\cong \angle A''C''B''$ in the inclined plane problem?
Someone have any other idea?